# Continuous set of eigenvalues in matrix representation?

1. Dec 23, 2014

Let's see if I have this straight: Observables are represented by Hermitian operators, which can be, for some appropriate base, represented in matrix form with the eigenvalues forming the diagonal. Sounds nice until I consider observables with continuous spectra. How do you get something like that into matrix form?

2. Dec 23, 2014

### vanhees71

The representation of operators in the eigenbasis is not a matrix, if there are continuous spectra. Then you have distributions. Take the position representation of the position operator in one-dimensional space. Its spectrum is entirely continous (entire $\mathbb{R}$). It's position representation is
$X(x_1,x_2)=\langle x_1|\hat{x} x_2 \rangle=x_2 \delta(x_1-x_2).$
That's not a matrix but a distribution. The action of the operator on a vector in such a case is not given by a usual matrix-vector product but by a convolution integral. Of course, in position representation that's very simple
$\langle x|\hat{x} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' X(x,x') \langle x'|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' X(x,x') \psi(x')=x \psi(x)$
as it must be due to the direct evaluation
$\langle x|\hat{x} \psi \rangle = \langle \hat{x} x|\psi \rangle=x \langle x |\psi \rangle=x \psi(x).$

3. Dec 23, 2014